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Chapter 5 Divide and Conquer Slides by Kevin Wayne. Copyright © 2005 Pearson-Addison Wesley. All rights reserved. 1 Divide-and-Conquer Divide-and-conquer. Break up problem into several parts. Solve each part recursively. Combine solutions to sub-problems into overall solution. Most common usage. Break up problem of size n into two equal parts of size ½n. Solve two parts recursively. Combine two solutions into overall solution in linear time. Consequence. Brute force: n2. Divide-and-conquer: n log n. Divide et impera. Veni, vidi, vici. - Julius Caesar 2 5.1 Mergesort Sorting Sorting. Given n elements, rearrange in ascending order. Obvious sorting applications. List files in a directory. Organize an MP3 library. List names in a phone book. Display Google PageRank results. Problems become easier once sorted. Find the median. Non-obvious sorting applications. Data compression. Computer graphics. Interval scheduling. Computational biology. Minimum spanning tree. Supply chain management. Simulate a system of particles. Book recommendations on Amazon. Load balancing on a 4 Mergesort Mergesort. Divide array into two halves. Recursively sort each half. Merge two halves to make sorted whole. Jon von Neumann (1945) A L G O R I T H M S A L G O R I T H M S divide O(1) A G L O R H I M S T sort 2T(n/2) merge O(n) A G H I L M O R S T 5 Merging Merging. Combine two pre-sorted lists into a sorted whole. How to merge efficiently? Linear number of comparisons. Use temporary array. A G A L G O H R H I M S T I Challenge for the bored. In-place merge. [Kronrud, 1969] using only a constant amount of extra storage 6 A Useful Recurrence Relation Def. T(n) = number of comparisons to mergesort an input of size n. Mergesort recurrence. 0 T(n) T n /2 solve left half T n /2 n solve right half merging if n 1 otherwise Solution. T(n) = O(n log2 n). Assorted proofs. We describe several ways to prove this recurrence. Initially we assume n is a power of 2 and replace with =. 7 Proof by Recursion Tree 0 T(n) 2T(n /2) n sorting both halves merging T(n) n 2(n/2) T(n/2) T(n/2) T(n/4) if n 1 otherwise T(n/4) T(n/4) T(n/4) log2n 4(n/4) ... 2k (n / 2k) T(n / 2k) ... T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) n/2 (2) n log2n 8 Proof by Telescoping Claim. If T(n) satisfies this recurrence, then T(n) = n log2 n. assumes n is a power of 2 0 T(n) 2T(n /2) n sorting both halves merging Pf. For n> 1: T(n) n if n 1 otherwise 2T(n /2) n 1 T(n /2) n /2 1 T(n / 4) n/4 1 1 T(n /n) n /n 1 1 log 2 n log2 n 9 Proof by Induction Claim. If T(n) satisfies this recurrence, then T(n) = n log2 n. assumes n is a power of 2 0 T(n) 2T(n /2) n sorting both halves merging if n 1 otherwise Pf. (by induction on n) Base case: n = 1. Inductive hypothesis: T(n) = n log2 n. Goal: show that T(2n) = 2n log2 (2n). T(2n) 2T(n) 2n 2n log2 n 2n 2nlog2 (2n) 1 2n 2n log2 (2n) 10 Analysis of Mergesort Recurrence Claim. If T(n) satisfies the following recurrence, then T(n) n lg n. 0 T(n) T n /2 T n /2 n merging solve right half solve left half log2n if n 1 otherwise Pf. (by induction on n) Base case: n = 1. Define n1 = n / 2 , n2 = n / 2. Induction step: assume true for 1, 2, ... , n–1. T(n) T(n1 ) T(n2 ) n n1lg n1 n2 lg n2 n n1 lg n2 n2 lg n2 n n lg n2 n n( lg n1 ) n n lg n n2 n /2 2 2 lg n lg n /2 /2 lg n2 lg n 1 11 5.3 Counting Inversions Counting Inversions Music site tries to match your song preferences with others. You rank n songs. Music site consults database to find people with similar tastes. Similarity metric: number of inversions between two rankings. My rank: 1, 2, …, n. Your rank: a1, a2, …, an. Songs i and j inverted if i < j, but ai > aj. Songs A B C D E Me 1 2 3 4 5 You 1 3 4 2 5 Inversions 3-2, 4-2 Brute force: check all (n2) pairs i and j. 13 Applications Applications. Voting theory. Collaborative filtering. Measuring the "sortedness" of an array. Sensitivity analysis of Google's ranking function. Rank aggregation for meta-searching on the Web. Nonparametric statistics (e.g., Kendall's Tau distance). 14 Counting Inversions: Divide-and-Conquer Divide-and-conquer. 1 5 4 8 10 2 6 9 12 11 3 7 15 Counting Inversions: Divide-and-Conquer Divide-and-conquer. Divide: separate list into two pieces. 1 1 5 5 4 4 8 8 10 10 2 2 6 6 9 9 12 12 11 11 3 3 7 Divide: O(1). 7 16 Counting Inversions: Divide-and-Conquer Divide-and-conquer. Divide: separate list into two pieces. Conquer: recursively count inversions in each half. 1 1 5 5 4 4 8 8 10 10 5 blue-blue inversions 5-4, 5-2, 4-2, 8-2, 10-2 2 2 6 6 9 9 12 12 11 11 3 3 7 7 Divide: O(1). Conquer: 2T(n / 2) 8 green-green inversions 6-3, 9-3, 9-7, 12-3, 12-7, 12-11, 11-3, 11-7 17 Counting Inversions: Divide-and-Conquer Divide-and-conquer. Divide: separate list into two pieces. Conquer: recursively count inversions in each half. Combine: count inversions where ai and aj are in different halves, and return sum of three quantities. 1 1 5 5 4 4 8 8 10 10 2 2 6 6 5 blue-blue inversions 9 9 12 12 11 11 3 3 7 7 Divide: O(1). Conquer: 2T(n / 2) 8 green-green inversions 9 blue-green inversions 5-3, 4-3, 8-6, 8-3, 8-7, 10-6, 10-9, 10-3, 10-7 Combine: ??? Total = 5 + 8 + 9 = 22. 18 Counting Inversions: Combine Combine: count blue-green inversions Assume each half is sorted. Count inversions where ai and aj are in different halves. Merge two sorted halves into sorted whole. to maintain sorted invariant 3 7 10 14 18 19 2 11 16 17 23 25 6 3 2 2 0 0 13 blue-green inversions: 6 + 3 + 2 + 2 + 0 + 0 2 3 7 10 11 14 16 17 18 19 Count: O(n) 23 25 Merge: O(n) T(n) T n/2 T n/2 O(n) T(n) O(nlog n) 19 Counting Inversions: Implementation Pre-condition. [Merge-and-Count] A and B are sorted. Post-condition. [Sort-and-Count] L is sorted. Sort-and-Count(L) { if list L has one element return 0 and the list L Divide the list into two halves A and B (rA, A) Sort-and-Count(A) (rB, B) Sort-and-Count(B) (rB, L) Merge-and-Count(A, B) } return r = rA + rB + r and the sorted list L 20 5.4 Closest Pair of Points Closest Pair of Points Closest pair. Given n points in the plane, find a pair with smallest Euclidean distance between them. Fundamental geometric primitive. Graphics, computer vision, geographic information systems, molecular modeling, air traffic control. Special case of nearest neighbor, Euclidean MST, Voronoi. fast closest pair inspired fast algorithms for these problems Brute force. Check all pairs of points p and q with (n2) comparisons. 1-D version. O(n log n) easy if points are on a line. Assumption. No two points have same x coordinate. to make presentation cleaner 22 Closest Pair of Points: First Attempt Divide. Sub-divide region into 4 quadrants. L 23 Closest Pair of Points: First Attempt Divide. Sub-divide region into 4 quadrants. Obstacle. Impossible to ensure n/4 points in each piece. L 24 Closest Pair of Points Algorithm. Divide: draw vertical line L so that roughly ½n points on each side. L 25 Closest Pair of Points Algorithm. Divide: draw vertical line L so that roughly ½n points on each side. Conquer: find closest pair in each side recursively. L 21 12 26 Closest Pair of Points Algorithm. Divide: draw vertical line L so that roughly ½n points on each side. Conquer: find closest pair in each side recursively. seems like (n2) Combine: find closest pair with one point in each side. Return best of 3 solutions. L 8 21 12 27 Closest Pair of Points Find closest pair with one point in each side, assuming that distance < . L 21 12 = min(12, 21) 28 Closest Pair of Points Find closest pair with one point in each side, assuming that distance < . Observation: only need to consider points within of line L. L 21 = min(12, 21) 12 29 Closest Pair of Points Find closest pair with one point in each side, assuming that distance < . Observation: only need to consider points within of line L. Sort points in 2-strip by their y coordinate. L 7 6 4 12 5 21 = min(12, 21) 3 2 1 30 Closest Pair of Points Find closest pair with one point in each side, assuming that distance < . Observation: only need to consider points within of line L. Sort points in 2-strip by their y coordinate. Only check distances of those within 11 positions in sorted list! L 7 6 4 12 5 21 = min(12, 21) 3 2 1 31 Closest Pair of Points Def. Let si be the point in the 2-strip, with the ith smallest y-coordinate. Claim. If |i – j| 12, then the distance between si and sj is at least . Pf. No two points lie in same ½-by-½ box. Two points at least 2 rows apart 2 rows have distance 2(½). ▪ j 39 31 ½ Fact. Still true if we replace 12 with 7. i ½ 30 29 28 27 ½ 26 25 32 Closest Pair Algorithm Closest-Pair(p1, …, pn) { Compute separation line L such that half the points are on one side and half on the other side. 1 = Closest-Pair(left half) 2 = Closest-Pair(right half) = min(1, 2) O(n log n) 2T(n / 2) Delete all points further than from separation line L O(n) Sort remaining points by y-coordinate. O(n log n) Scan points in y-order and compare distance between each point and next 11 neighbors. If any of these distances is less than , update . O(n) return . } 33 Closest Pair of Points: Analysis Running time. T(n) 2T n/2 O(n log n) T(n) O(n log 2 n) Q. Can we achieve O(n log n)? A. Yes. Don't sort points in strip from scratch each time. Each recursive returns two lists: all points sorted by y coordinate, and all points sorted by x coordinate. Sort by merging two pre-sorted lists. T(n) 2T n/2 O(n) T(n) O(n log n) 34 5.5 Integer Multiplication Integer Arithmetic Add. Given two n-digit integers a and b, compute a + b. O(n) bit operations. Multiply. Given two n-digit integers a and b, compute a × b. Brute force solution: (n2) bit operations. 1 1 0 1 0 1 0 1 * 0 1 1 1 1 1 0 1 1 1 0 1 0 1 0 1 0 Multiply 0 0 0 0 0 0 0 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 1 1 1 1 1 0 1 1 1 0 1 0 1 0 1 + 0 1 1 1 1 1 0 1 1 0 1 0 1 0 0 1 0 Add 1 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 36 Divide-and-Conquer Multiplication: Warmup To multiply two n-digit integers: Multiply four ½n-digit integers. Add two ½n-digit integers, and shift to obtain result. x 2 n / 2 x1 x0 y 2 n / 2 y1 y0 xy 2 n / 2 x1 x0 2 n / 2 y1 y0 2 n x1 y1 2 n / 2 x1 y0 x0 y1 x0 y0 T(n) 4T n /2 (n) recursive calls T(n) (n 2 ) add, shift assumes n is a power of 2 37 Karatsuba Multiplication To multiply two n-digit integers: Add two ½n digit integers. Multiply three ½n-digit integers. Add, subtract, and shift ½n-digit integers to obtain result. x 2 n / 2 x1 y 2 n / 2 y1 xy 2 n x1 y1 2 n x1 y1 A x0 y0 2 n / 2 x1 y0 x0 y1 x0 y0 2 n / 2 (x1 x0 ) (y1 y0 ) x1 y1 x0 y0 x0 y0 B A C C Theorem. [Karatsuba-Ofman, 1962] Can multiply two n-digit integers in O(n1.585) bit operations. T(n) T n /2 T n /2 T 1 n /2 recursive calls T(n) O(n log 2 3 (n) add, subtract, shift ) O(n1.585 ) 38 Karatsuba: Recursion Tree 0 if n 1 T(n) 3T(n/2) n otherwise log 2 n T(n) n k0 3 k 2 23 1 log 2 n 3 1 2 1 3nlog 2 3 2 n T(n) T(n/2) T(n/2) T(n/2) 3(n/2) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) 9(n/4) ... ... 3k (n / 2k) T(n / 2k) ... T(2) T(2) T(2) T(2) ... T(2) T(2) T(2) T(2) 3 lg n (2) 39 Matrix Multiplication Matrix Multiplication Matrix multiplication. Given two n-by-n matrices A and B, compute C = AB. c11 c12 c21 c22 c n1 cn2 n cij a ik bkj k 1 a11 a12 c1n c2n a a 22 21 a cnn n1 an2 b11 b12 a1n a 2n b b 21 22 b b ann n1 n2 b1n b2n bnn Brute force. (n3) arithmetic operations. Fundamental question. Can we improve upon brute force? 41 Matrix Multiplication: Warmup Divide-and-conquer. Divide: partition A and B into ½n-by-½n blocks. Conquer: multiply 8 ½n-by-½n recursively. Combine: add appropriate products using 4 matrix additions. C11 C12 C C 21 22 A11 A21 A12 A22 B11 B21 T(n) 8T n /2 recursive calls B12 B22 (n2 ) C11 C12 C21 C22 A11 B11 A12 B21 A11 B12 A12 B22 A21 B11 A22 B21 A21 B12 A22 B22 T(n) (n 3 ) add, form submatrices 42 Matrix Multiplication: Key Idea Key idea. multiply 2-by-2 block matrices with only 7 multiplications. C11 C21 C12 A11 C22 A21 C11 C12 C21 C22 A12 A22 B11 B21 B12 B22 P5 P4 P2 P6 P1 P2 P3 P4 P5 P1 P3 P7 P1 P2 P3 P4 P5 P6 P7 A11 ( B12 B22 ) ( A11 A12 ) B22 ( A21 A22 ) B11 A22 ( B21 B11 ) ( A11 A22 ) ( B11 B22 ) ( A12 A22 ) ( B21 B22 ) ( A11 A21 ) ( B11 B12 ) 7 multiplications. 18 = 10 + 8 additions (or subtractions). 43 Fast Matrix Multiplication Fast matrix multiplication. (Strassen, 1969) Divide: partition A and B into ½n-by-½n blocks. Compute: 14 ½n-by-½n matrices via 10 matrix additions. Conquer: multiply 7 ½n-by-½n matrices recursively. Combine: 7 products into 4 terms using 8 matrix additions. Analysis. Assume n is a power of 2. T(n) = # arithmetic operations. T(n) 7T n /2 recursive calls (n 2 ) T(n) (n log 2 7 ) O(n 2.81 ) add, subtract 44 Fast Matrix Multiplication in Practice Implementation issues. Sparsity. Caching effects. Numerical stability. Odd matrix dimensions. Crossover to classical algorithm around n = 128. Common misperception: "Strassen is only a theoretical curiosity." Advanced Computation Group at Apple Computer reports 8x speedup on G4 Velocity Engine when n ~ 2,500. Range of instances where it's useful is a subject of controversy. Remark. Can "Strassenize" Ax=b, determinant, eigenvalues, and other matrix ops. 45 Fast Matrix Multiplication in Theory Q. Multiply two 2-by-2 matrices with only 7 scalar multiplications? A. Yes! [Strassen, 1969] (n log 2 7 ) O(n 2.81 ) Q. Multiply two 2-by-2 matrices with only 6 scalar multiplications? A. Impossible. [Hopcroft and Kerr, 1971] log 2 6 2.59 (n ) O(n ) Q. Two 3-by-3 matrices with only 21 scalar multiplications? A. Also impossible. (n log 3 21 ) O(n 2.77 ) Q. Two 70-by-70 matrices with only 143,640 scalar multiplications? A. Yes! [Pan, 1980] log 70 143640 2.80 (n Decimal wars. December, 1979: O(n2.521813). January, 1980: O(n2.521801). ) O(n ) 46 Fast Matrix Multiplication in Theory Best known. O(n2.376) [Coppersmith-Winograd, 1987.] Conjecture. O(n2+) for any > 0. Caveat. Theoretical improvements to Strassen are progressively less practical. 47